Level of Each node in a Tree from source node (using BFS) (original) (raw)

Last Updated : 1 Aug, 2022

Given a tree with v vertices, find the level of each node in a tree from the source node.

Examples:

Input :

Output : Node Level 0 0 1 1 2 1 3 2 4 2 5 2 6 2 7 3

Explanation :

Input:

Output : Node Level 0 0 1 1 2 1 3 2 4 2 Explanation:

Approach:
BFS(Breadth-First Search) is a graph traversal technique where a node and its neighbours are visited first and then the neighbours of neighbours. In simple terms, it traverses level-wise from the source. First, it traverses level 1 nodes (direct neighbours of source node) and then level 2 nodes (neighbours of source node) and so on. The BFS can be used to determine the level of each node from a given source node.

Algorithm:

1. Create the tree, a queue to store the nodes and insert the root or starting node in the queue. Create an extra array level of size v (number of vertices) and create a visited array.
2. Run a loop while size of queue is greater than 0.
3. Mark the current node as visited.
4. Pop one node from the queue and insert its childrens (if present) and update the size of the inserted node as level[child] = level[node] + 1.
5. Print all the node and its level.

Implementation:

C++ `

// CPP Program to determine level of each node // and print level #include <bits/stdc++.h> using namespace std;

// function to determine level of each node starting // from x using BFS void printLevels(vector graph[], int V, int x) { // array to store level of each node int level[V]; bool marked[V];

// create a queue
queue<int> que;

// enqueue element x
que.push(x);

// initialize level of source node to 0
level[x] = 0;

// marked it as visited
marked[x] = true;

// do until queue is empty
while (!que.empty()) {

    // get the first element of queue
    x = que.front();

    // dequeue element
    que.pop();

    // traverse neighbors of node x
    for (int i = 0; i < graph[x].size(); i++) {
        // b is neighbor of node x
        int b = graph[x][i];

        // if b is not marked already
        if (!marked[b]) {

            // enqueue b in queue
            que.push(b);

            // level of b is level of x + 1
            level[b] = level[x] + 1;

            // mark b
            marked[b] = true;
        }
    }
}

// display all nodes and their levels
cout << "Nodes"
     << "    "
     << "Level" << endl;
for (int i = 0; i < V; i++)
    cout << " " << i << "   -->   " << level[i] << endl;

}

// Driver Code int main() { // adjacency graph for tree int V = 8; vector graph[V];

graph[0].push_back(1);
graph[0].push_back(2);
graph[1].push_back(3);
graph[1].push_back(4);
graph[1].push_back(5);
graph[2].push_back(5);
graph[2].push_back(6);
graph[6].push_back(7);

// call levels function with source as 0
printLevels(graph, V, 0);

return 0;

}

Java

// Java Program to determine level of each node // and print level import java.util.*;

class GFG {

// function to determine level of each node starting
// from x using BFS
static void printLevels(Vector<Vector<Integer> > graph,
                        int V, int x)
{
    // array to store level of each node
    int level[] = new int[V];
    boolean marked[] = new boolean[V];

    // create a queue
    Queue<Integer> que = new LinkedList<Integer>();

    // enqueue element x
    que.add(x);

    // initialize level of source node to 0
    level[x] = 0;

    // marked it as visited
    marked[x] = true;

    // do until queue is empty
    while (que.size() > 0) {

        // get the first element of queue
        x = que.peek();

        // dequeue element
        que.remove();

        // traverse neighbors of node x
        for (int i = 0; i < graph.get(x).size(); i++) {
            // b is neighbor of node x
            int b = graph.get(x).get(i);

            // if b is not marked already
            if (!marked[b]) {

                // enqueue b in queue
                que.add(b);

                // level of b is level of x + 1
                level[b] = level[x] + 1;

                // mark b
                marked[b] = true;
            }
        }
    }

    // display all nodes and their levels
    System.out.println("Nodes"
                       + " "
                       + "Level");
    for (int i = 0; i < V; i++)
        System.out.println(" " + i + " --> "
                           + level[i]);
}

// Driver Code
public static void main(String args[])
{
    // adjacency graph for tree
    int V = 8;
    Vector<Vector<Integer> > graph
        = new Vector<Vector<Integer> >();

    for (int i = 0; i < V + 1; i++)
        graph.add(new Vector<Integer>());

    graph.get(0).add(1);
    graph.get(0).add(2);
    graph.get(1).add(3);
    graph.get(1).add(4);
    graph.get(1).add(5);
    graph.get(2).add(5);
    graph.get(2).add(6);
    graph.get(6).add(7);

    // call levels function with source as 0
    printLevels(graph, V, 0);
}

}

// This code is contributed by Arnab Kundu

Python3

Python3 Program to determine level

of each node and print level

import queue

function to determine level of

each node starting from x using BFS

def printLevels(graph, V, x):

# array to store level of each node
level = [None] * V
marked = [False] * V

# create a queue
que = queue.Queue()

# enqueue element x
que.put(x)

# initialize level of source
# node to 0
level[x] = 0

# marked it as visited
marked[x] = True

# do until queue is empty
while (not que.empty()):

    # get the first element of queue
    x = que.get()

    # traverse neighbors of node x
    for i in range(len(graph[x])):

        # b is neighbor of node x
        b = graph[x][i]

        # if b is not marked already
        if (not marked[b]):

            # enqueue b in queue
            que.put(b)

            # level of b is level of x + 1
            level[b] = level[x] + 1

            # mark b
            marked[b] = True

# display all nodes and their levels
print("Nodes", " ", "Level")
for i in range(V):
    print(" ", i,  " --> ", level[i])

Driver Code

if name == 'main':

# adjacency graph for tree
V = 8
graph = [[] for i in range(V)]

graph[0].append(1)
graph[0].append(2)
graph[1].append(3)
graph[1].append(4)
graph[1].append(5)
graph[2].append(5)
graph[2].append(6)
graph[6].append(7)

# call levels function with source as 0
printLevels(graph, V, 0)

This code is contributed by PranchalK

C#

// C# Program to determine level of each node // and print level using System; using System.Collections.Generic;

class GFG {

// function to determine level of each node starting
// from x using BFS
static void printLevels(List<List<int> > graph, int V,
                        int x)
{
    // array to store level of each node
    int[] level = new int[V];
    Boolean[] marked = new Boolean[V];

    // create a queue
    Queue<int> que = new Queue<int>();

    // enqueue element x
    que.Enqueue(x);

    // initialize level of source node to 0
    level[x] = 0;

    // marked it as visited
    marked[x] = true;

    // do until queue is empty
    while (que.Count > 0) {

        // get the first element of queue
        x = que.Peek();

        // dequeue element
        que.Dequeue();

        // traverse neighbors of node x
        for (int i = 0; i < graph[x].Count; i++) {
            // b is neighbor of node x
            int b = graph[x][i];

            // if b is not marked already
            if (!marked[b]) {

                // enqueue b in queue
                que.Enqueue(b);

                // level of b is level of x + 1
                level[b] = level[x] + 1;

                // mark b
                marked[b] = true;
            }
        }
    }

    // display all nodes and their levels
    Console.WriteLine("Nodes"
                      + " "
                      + "Level");
    for (int i = 0; i < V; i++)
        Console.WriteLine(" " + i + " --> " + level[i]);
}

// Driver Code
public static void Main(String[] args)
{
    // adjacency graph for tree
    int V = 8;
    List<List<int> > graph = new List<List<int> >();

    for (int i = 0; i < V + 1; i++)
        graph.Add(new List<int>());

    graph[0].Add(1);
    graph[0].Add(2);
    graph[1].Add(3);
    graph[1].Add(4);
    graph[1].Add(5);
    graph[2].Add(5);
    graph[2].Add(6);
    graph[6].Add(7);

    // call levels function with source as 0
    printLevels(graph, V, 0);
}

}

// This code is contributed by Princi Singh

JavaScript

`

Output

Nodes Level 0 --> 0 1 --> 1 2 --> 1 3 --> 2 4 --> 2 5 --> 2 6 --> 2 7 --> 3

Complexity Analysis:

• Time Complexity: O(n).
  In BFS traversal every node is visited only once, so Time Complexity is O(n).
• Space Complexity: O(n).
  The space is required to store the nodes in a queue.